Question

If x is real, the expression $\frac{x+2}{(2x^2+3x+6)}$ takes all value in the interval

• A. $(\frac{1}{13},\frac{1}{3})$
• B. $(-\frac{1}{13},\frac{1}{3})$
• C. $(\frac{1}{3},\frac{1}{13})$
• D. None of these

Answer 1

Answer: (B)

$(-\frac{1}{13},\frac{1}{3})$

$Given,\frac{x+2}{(2x^2+3x+6)}if:y=\frac{x+2}{(2x^2+3x+6)}\Rightarrow 2y{x}^2+3yx+6y=x+2\Rightarrow 2y{x}^2+3yx+6y-x-2=0\Rightarrow 2y{x}^2+3yx+6y-x-2=0\Rightarrow 2y{x}^2+(3y-1)x+6y-2=0here,xisrealno.andD\ge 0\Rightarrow (3y-1{)}^2-4\left(2y\right)(6y-2)\ge 0\Rightarrow 9y^2+1-6y-8y(6y-2)\ge 0\Rightarrow 9y^2+1-6y-48y^2+16y)\ge 0\Rightarrow -39y^2+10y+1\ge 0\Rightarrow 39y^2-10y-1\le 0\Rightarrow 39y^2-13y+3y-1\le 0\Rightarrow 13y(3y-1)+1(3y-1)\le 0\Rightarrow (3y-1\left)\right(13y+1)\le 0\therefore y=\left[\frac{-1}{13},\frac{1}{3}\right]so,thatthecorrectoptionisB.$